Electricity Name_____________

Question 1.

(a) Describe the motion of the free charge carriers in a current-carrying wire. (3 marks)

charge carriers = (free) electrons (1mark)collisions with atoms (or ions) (1mark)negative electron drift towards the positive terminal (1mark)OR if there is a potential difference

(b) A current of 2.0mA flows for 5.0 minutes,Calculate1. The total charge which passes any point in the wire.

Q=It = 2x10-3 x 5 x 60

Charge = 0.6 C (2 marks)

2. The number of electrons which pass the point during this time.

n= 0.6/1.6 x 10-19

Number=3.75 x 1018 2 marks)(e = 1.60 x 10-19 C)

Question 2.There are four different ways of connecting together three identical resistors, each of resistance 20. Sketch and three of the possible methods of connection and work out the effective resistance of each combination. (4 marks)

(1) Resistance = ______________

(2) Resistance = ______________

(3) Resistance = ______________

60

13.3

6.7

30

Question 3.

A battery with a finite internal resistance has an electromotive force (e.m.f) of 20 V.

(a) Define electromotive force of a battery (2 marks)

Terminal potential difference when no current is drawn from battery (or) energy converted to electrical form per unit charge passing through the battery.

(b) The battery drives a current of 0.50A through a resistor of resistance 36 connected across its terminals.

(i) Calculate the internal resistance of the battery. (3 marks)ALWAYS DRAW THE DIAGRAM WITH THE INTERNAL RESISTANCE.

Internal resistance = 4

(ii) The current passes for a time of 6.0 minutes.Calculate

(1) The electrical energy used up in the 36 resistor in this time,

P =VI E = Pt = VIt =I2Rt=(0.5)2 x 36 x 6 x60

Energy = ____3240______J (2 marks)

(2) The total electrical energy used up in the complete circuit (including the battery) in this time.

E = I2Rt =(0.5)2 x 40x 6 x60

Energy = ____3600______ J (2 marks)

36

r

20VV=IR I=V/R = 30/(36+r)=0.5

0.5(36+r)=20

r=40-36=4

Question 3.

(a) Electrical devices which obey Ohm’s law are described as ohmic. Explain what is meant by ohmic behaviour. (2 marks)

P.d (across device) is proportional to current OR V I (1 mark)OR V/I = constant

when at constant temperature (1 mark)

(a) Describe an experiment to determine the current voltage characteristics of a tungsten filament lamp. Your answer should include a circuit diagram, an account of the procedure , and a sketch graph of the results you would expect to obtain. (6 marks)

V

I

0

(c) Explain why the graph you have sketched in (b) has this particular shape. (3 marks)

For large current, filament temperature is high. Resistance of metal increases with temperature as ions vibrate more due to increased K.E and there are therefore more collisions with the electrons. R=V/I

OR (R = 1/slope and is changing due to ….)

Quality of written communication (1 mark)

(1 mark)A

V(3 marks)

Read current and voltage, adjust rheostat and repeat for different currents and voltages.(2 marks)

Question 5.

A p.d of 2.0V is applied across the ends of a 1.5m length of copper wire of diameter 0.20mmCalculate (a) the resistance of the wire(b) the current in the wire.

For copper, number of free electrons per m3 = 1.0x1029;resistivity = 1.7x10-8m (3 marks)

P = VI (watts) = V2/R

R = V2/P …………(1)

From 1 & 2:

R = 4L =V2

d2 P

d2 = 4LP = 4 x 1.1x10-6 x 2.4 x 400 V2 (240)2

d= 1.5 x 10-4 mm

Question 6.

The heating element in a hair dryer dissipates energy at a rate of 400W when it is connected to a 240V supply. If the element is made of nichrome wire of total length 2.4m, calculate the diameter of the wire used. (Resistivity of nichrome = 1.1x10-6 m) (3marks)

R = L A = L A R

r2 = (d/2)2 = d2 = L 4 R

R = 4L ………….(2) d2

(a) R = L = (1.7x10-8)(1.5) = 4(1.7x10-8)(1.5) = 0.81 A r2 d2

(b) Current density = current per cross sect areaV=IR I= V/R = 2/0.81 = 2.5A

therefore current density = I/A = 2.5/r2 = 10/ d2 = 7.9 x107 Am-2

Question 7For the circuit shownCalculate (5 marks)(a) the total resistance of the circuit(b) the currents and p.d.s shown

24I1

I1.2

6I2

15V

V1

V2

(a) Parallel resistors R=(24x6)/(24+6) = 4.8

Rtotal = 1.2 + 4.8 = 6

(b) V=IR I= V/R = 15/6 = 2.5A

I = 2.5 A and V1 = 2.5 x 1.2 = 3V

Resistor ratio = 6:24 = 1:4, therefore

1/5 current through 24 resitor = 1/5 x 2.5 = 0.5A

I1=0.5A V2 = 0.5 x 24 =12V

4/5 current through 6 resistor = 4/5 x 2.5 = 2 A

I2=2A and V2 = 2 x 6 =12V

Question 8.

For the circuit shown

Calculate (5 marks)

(a) the total resistance of the circuit

(b) the total current (I)

(c) the branch currents (I1 and I2)

(d) the p.d. across resistor B

(e) the p.d. across resistor D

10V

50k

50k50k

100k

I

I1

I2

A

C D

B

(a) Parallel resistors R=(150x100)/(250) = 60k = 6x104

(b) V=IR I= V/R = 10/6x104 = 1.7x10-4A

(c) Resistor ratio = 100:150 = 10:15 = 2:3 , therefore 2/5 current through (A and B) = 2/5 x 1.7x10-4 = 6.7x10-5A = I1

3/5 current through (C and D) = 3/5 x 1.7x10-4 = 1.0x10-4A = I2

(d) p.d across B = Vb = I1Rb = 6.7x10-5x 100x103=6.7V

(e) p.d across D = Vd = I2Rd = 1.0x10-4x 50x103=5.1V